a: \(65\cdot48+35\cdot48+12\cdot58-12\cdot38\)

\(=48\left(65+35\right)+12\left(58-38\right)\)

\(=48\cdot100+12\cdot20\)

\(=4800+240=5040\)

b: \(2018^2+\left\{15^2:\left[175-\left(2^3\cdot5^2-6\cdot25\right)\right]\right\}\)

\(=1+\left\{\dfrac{225}{175-\left(8\cdot25-150\right)}\right\}\)

\(=1+\dfrac{225}{175-\left(200-150\right)}\)

\(=1+\dfrac{225}{175-50}=1+\dfrac{225}{125}=1+\dfrac{5}{3}=\dfrac{8}{3}\)

a,  15 . { 32 : [ 6 - 5 + 5 ( 9 : 3 ) ] + 3 } - 2018 0

= 15.{32:[1+15]+3}–1

= 15.5–1

= 74

b,  25 . { 2 7 : [ 12 - 4 + 2 2 . 16 : 2 3 ] - 2 4 }

= 25.{128:[8+4.2]–16}

= 25.24

= 600

c,  2019 . { 101 - 1000 : [ 2 2 . 2 3 + 5 6 : 5 3 - 6 2 : 11 - 2018 0 ] }

= 2019.{101–1000:[(32+125–36):11–1]}

= 2019.{101–1000:[121:11–1]}

= 2019.{101–1000:10}

= 2019.1

= 2019

x - (40 - 67) = 2018⁰

x - (-27) = 1

x + 27 = 1

x = 1 - 27

x = -26

\(x\) - (40 - 67) = 2018⁰

\(x\) + (67 - 40) = 1

\(x\) + 27 = 1

\(x\)          = 1 - 27

\(x\)          = - 26 

\(\left(1+3+5+7+...+65+67\right)\cdot\left(125\cdot1001-125125\right)+10\)

\(=\left(1+3+5+...+65+67\right)\cdot\left(125\cdot1000+125-125125\right)+10\)

\(=\left(1+3+5+...+65+67\right)\cdot\left(125000+125-125125\right)+10\)

\(=\left(1+3+5+...+65+67\right)\cdot\left(125125-125125\right)+10\)

\(=\left(1+3+5+...+65+67\right)\cdot0+10\)

\(=0+10\)

\(=10\)

Hơi dài dòng nhưng đầy đủ

a, 36:(x–5) =  2 2

(x–5) = 9

x = 14

b, [3.(70–x)+5]:2 = 46

[3.(70–x)+5] = 92

70–x = 29

x = 41

c, 450:[41–(2x–5)] =  3 2 .5

41–(2x–5) = 10

2x–5 = 31

2x = 36

x = 18

d, 230+[ 2 4 +(x–5)] = 315. 2018 0

16+(x–5) = 315–230

x–5 = 85–16

x = 69+5

x = 74

e,  2 x + 2 x + 1  = 48

2 x .(2+1) = 48

2 x = 16 =  2 4

x = 4

f,  3 x + 2 + 3 x  = 2430

3 x . 3 2 + 1 = 2430

3 x = 2430:10 = 243 =  3 5

x = 5

\(=-\dfrac{8}{5}\left|-\dfrac{3}{4}\right|+1=-\dfrac{8}{5}\cdot\dfrac{3}{4}+1=-\dfrac{6}{5}+1=-\dfrac{1}{5}\)

a: \(x-\dfrac{10}{3}=\dfrac{7}{15}\cdot\dfrac{3}{5}\)

=>\(x-\dfrac{10}{3}=\dfrac{21}{75}=\dfrac{7}{25}\)

=>\(x=\dfrac{7}{25}+\dfrac{10}{3}=\dfrac{21+250}{75}=\dfrac{271}{75}\)

b: \(x+\dfrac{3}{22}=\dfrac{27}{121}\cdot\dfrac{9}{11}\)

=>\(x+\dfrac{3}{22}=\dfrac{243}{1331}\)

=>\(x=\dfrac{243}{1331}-\dfrac{3}{22}=\dfrac{123}{2662}\)

c: \(\dfrac{8}{23}\cdot\dfrac{46}{24}-x=\dfrac{1}{3}\)

=>\(\dfrac{8}{24}\cdot\dfrac{46}{23}-x=\dfrac{1}{3}\)

=>\(\dfrac{2}{3}-x=\dfrac{1}{3}\)

=>\(x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

d: \(1-x=\dfrac{49}{65}\cdot\dfrac{5}{7}\)

=>\(1-x=\dfrac{49}{7}\cdot\dfrac{5}{65}=\dfrac{7}{13}\)

=>\(x=1-\dfrac{7}{13}=\dfrac{6}{13}\)

a) \(-65-\left(x+15\right)+105=0\)

\(-65-\left(x+15\right)=-105\)

\(x+15=-65-\left(-105\right)\)

\(x+15=40\)

\(x=25\)

b) |-6| + (-9) - (x+1) = 7

-3- (x+1) =7

       x+1  = -10

       x      = -11

c)\(\left|5-x\right|+\left(-25+7\right)=-3-\left(-10\right)\)

    \(\left|5-x\right|+\left(-18\right)=7\)

    \(\left|5-x\right|=25\)

\(\Rightarrow\orbr{\begin{cases}5-x=25\\5-x=-25\end{cases}\Rightarrow\orbr{\begin{cases}x=-20\\x=30\end{cases}}}\)

d) \(28-\left|x+6\right|+\left(-2\right)=0\)

\(28-\left|x+6\right|=2\)

\(\left|x+6\right|=26\)

\(\Rightarrow\orbr{\begin{cases}x+6=26\\x+6=-26\end{cases}\Rightarrow\orbr{\begin{cases}x=20\\x=-32\end{cases}}}\)

e) \(x-\left(13-15\right)=5+\left(10-x\right)-\left(-1\right)\)

\(x+2=5+10-x+1\)

\(x+2=16-x\)

\(x+x=16-2\)

\(2x=14\)

\(x=7\)

f) \(-120-\left(x-5\right)=125\)

\(x-5=-120-125\)

\(x-5=-245\)

\(x=-240\)

g) \(10-\left(-5+2\right)+\left(-9\right)=\left(-20+7\right)-x\)

\(-16=-13-x\)

\(x=-13+16\)

\(x=3\)